Problem: Simplify the following expression and state the condition under which the simplification is valid. $q = \dfrac{3y^3 - 42y^2 + 147y}{-8y^2 - 24y + 560}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ q = \dfrac {3y(y^2 - 14y + 49)} {-8(y^2 + 3y - 70)} $ $ q = -\dfrac{3y}{8} \cdot \dfrac{y^2 - 14y + 49}{y^2 + 3y - 70} $ Next factor the numerator and denominator. $ q = - \dfrac{3y}{8} \cdot \dfrac{(y - 7)(y - 7)}{(y - 7)(y + 10)}$ Assuming $y \neq 7$ , we can cancel the $y - 7$ $ q = - \dfrac{3y}{8} \cdot \dfrac{y - 7}{y + 10}$ Therefore: $ q = \dfrac{ -3y(y - 7)}{ 8(y + 10)}$, $y \neq 7$